Problem: What is the value of $\dfrac{d}{dx}\left(\dfrac{9}{x^3}+\dfrac{1}{x}\right)$ at $x=-3$ ?
Answer: The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have the derivative, we can evaluate it at $x=-3$. Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}\dfrac{9}{x^3}+\dfrac{1}{x} \\\\ &=9x^{-3}+x^{-1} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(9x^{-3}+x^{-1} ) \\\\ &=9\dfrac{d}{dx}(x^{-3})+\dfrac{d}{dx}(x^{-1}) \\\\ &=9(-3x^{-4})+(-1)x^{-2} \\\\ &=-27x^{-4}-1x^{-2} \\\\ &=-\dfrac{27}{x^4}-\dfrac{1}{x^2} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\dfrac{9}{x^3}+\dfrac{1}{x}\right)=-\dfrac{27}{x^4}-\dfrac{1}{x^2}$. Let's evaluate it at $x=-3$ : $\begin{aligned} &\phantom{=}-\dfrac{27}{x^4}-\dfrac{1}{x^2} \\\\ &=-\dfrac{27}{(-3)^4}-\dfrac{1}{(-3)^2} \\\\ &=-\dfrac{27}{81}-\dfrac{1}{9} \\\\ &=-\dfrac{3}{9}-\dfrac{1}{9} \\\\ &=-\dfrac{4}{9} \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{9}{x^3}+\dfrac{1}{x}\right)$ at $x=-3$ is $-\dfrac{4}{9}$.